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In the proof presented in my textbook, it utilises the equation

$$(A-\lambda I) \operatorname{Adj}(A - \lambda I) = \det(A-\lambda I) \, I$$

And it stated that,

$$P(A-\lambda I) = \det(A-\lambda I)= a_0 + a_1\lambda + ... + a_n\lambda^n$$

and that

$$Q(A-\lambda I) = \operatorname{Adj}((A-\lambda I) = b_0 + b_1\lambda + ... + b_k\lambda^k$$

I have no problem understanding that $P(A-\lambda I)$ is the characteristic polynomial for matrix $A$. However, I have problem understanding 2 things,

  1. Why $\operatorname{Adj}(A-\lambda I)$ can be written into a polynomial?

  2. What does it mean to then multiply the above polynomial to matrix $(A-\lambda I)$?

Many thanks!

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  • $\begingroup$ You can answer your questions for $n=2$ yourself as follows. Write out all expressions explicitly for $A=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$. How do you define $Adj$? You can compute it explicitly for $A$, see here. $\endgroup$ – Dietrich Burde Mar 2 at 9:25
  • $\begingroup$ I mean if I just do the operation $\begin{pmatrix} d-\lambda & -b \cr -c & a-\lambda \end{pmatrix}$ $\begin{pmatrix} a-\lambda & b \cr c & d-\lambda \end{pmatrix}$. What I get is $\begin{pmatrix} (a-\lambda)(d- \lambda) - bc & 0 \cr 0 & (a-\lambda)(d-\lambda) + bc \end{pmatrix}$. Correct? $\endgroup$ – Klein_Bottle Mar 2 at 11:08
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    $\begingroup$ Adj$(A-\lambda I)$ can be thought of as a polynomial in $\lambda$ whose coefficients are matrices. I think Adj$(A-\lambda I)=B_0+B_1\lambda+\cdots +B_{n-1}\lambda^{n-1}$ is a better way of writing it. $\endgroup$ – Lord Shark the Unknown Mar 2 at 11:11
  • $\begingroup$ The equality in your question should be $\color{red}{(A-\lambda I)}\operatorname{Adj}(A-\lambda I)=\det(A-\lambda I)I$, not $\color{red}{A}\operatorname{Adj}(A-\lambda I)=\det(A-\lambda I)I$. $\endgroup$ – user1551 Mar 2 at 13:20
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I hope it's not too much trouble for me to talk in terms of endomorphisms rather than square matrices.

If we want to evaluate the polynomial $\text{det}(\phi - t I)$ at a matrix, it is natural to view it as an element of $\text{End}_{\mathbb{F}} (V) [t]$. The factorization $\text{det}(t I - \phi ) = \text{adj}(tI - \phi ) (tI - \phi)$ makes sense if we allow ourselves to think of $\text{adj}(tI - \phi)$ as a polynomial with endomorphisms as entries.

After we think of these as elements in $\text{End}_{\mathbb{F}} (V)[t]$, the product $\text{adj}(A - \lambda I) ( A - \lambda I)$ works just like any polynomial ring over a noncommutative ring:

$$\sum_{i = 1}^n \phi_i t^i \sum_{j = 1}^m \psi_j t^j = \sum_{i = 1}^{n + m} \sum_{j + k = i, 1 \leq j \leq n, 1 \leq k \leq m} \phi_j \circ \psi_k t^{i}$$

In fact, this language is convenient for the proof: we have an isomorphism $$\text{End}_{\mathbb{F}} (V) [t] \cong \text{End}_{\mathbb{F}} (V \otimes_{\mathbb{F}} \mathbb{F}[t])$$ In $\text{End}_{\mathbb{F}} (V \otimes_{\mathbb{F}} \mathbb{F}[t])$, we have a factorization $$ \text{det}(\mu_t - \phi) 1_V = \text{adj}(\mu_t - \phi) (\mu_t - \phi) $$ Where $\mu_t$ is multiplication by $t$. Under the isomorphism, this becomes a factorization $$ p(t) = f(t)(t - \phi) $$ where $p(t) \in \text{End}_{\mathbb{F}} (V)[t]$ is the characteristic polynomial.

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